How many grams of iron can be obtained from a 268g sample of iron 3 oxide?

Answer 1

Iron (III) oxide -> Iron + Oxygen

Steps to find the mass of iron obtained from 268g of iron (III) oxide:

Step 1: Write the balanced equation

2Fe2O3 -> 4Fe + 3O2

Step 2: Identify key reactants and products

2Fe2O3 -> 4Fe

Step 3: Write down the mole ratio

2 moles : 4 moles

Step 4: Change moles into gram formula mass

2((2x56)+(3x16)) : 4(1x56)

320g : 224g

Step 5: Find out how many grams of iron will be produced for every gram of iron (III) oxide

1 gram = 224/320 = 0.7g

Step 6: Find out how many grams of iron will be produced for 268 grams of iron (III) oxide

268 grams = (224/320)x268 = 187.6 grams

The mass of iron produced from 268 grams of iron (III) oxide is 187.6 grams.

Answer 2

To find the amount of iron in iron (III) oxide, we first need to calculate the molar mass of Fe2O3. The molar mass of Fe2O3 is 159.69 g/mol. By looking at the ratio of atoms in Fe2O3, we see that 2 moles of Fe are present in 1 mole of Fe2O3. By applying this ratio, we can find that 2 moles of Fe correspond to 159.69g, so 1 mole will be 79.85g. Next, we can calculate the amount of Fe in 268g of Fe2O3. By using stoichiometry, we determine that 1 mol of Fe is equal to 55.85 g. Multiplying this by the number of moles of Fe (79.85g/55.85g) present in the 268g sample of Fe2O3, we get that approximately 115g of Fe can be obtained from the 268g sample of Fe2O3.